# Ring of Integers as subring with most irreducibles

Let $L$ be a number field. Is it possible to define its ring of integers $R$ by saying it's the subring with (in a fuzzy sense) the "most" irreducibles?

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It seems you're looking for the universal property of normalization, which is the geometric version of "taking the integral closure". See mathoverflow.net/questions/46/… You might also want to replace "irreducibles" by "prime ideals" in your question. – François Brunault Feb 14 '12 at 13:27
You could make "most" less fuzzy by asking: Does the ring of integers satisfy the property that if there is a subring $S$ of $L$ in which $x\in S$ is irreducible, then $x\in R$ and is irreducible or a product of irreducibles. One problem is that, unless the class number is 1, the ring of integers is not a UFD. So irreducibles behave badly. This is why, historically, one uses prime ideals rather than irreducibles. We have unique factorization of prime ideals. – Pace Nielsen Feb 14 '12 at 22:03